Dear Mathlinkers, Hint of a proof : consider the center O of the circle and applie the Pascal theorem in a degenerated case... Sincerely Jean-Louis

Let $PG$ tangent to $(O)$, then $AG \bot PO$, so $CG\parallel PO$. As $BO = DO$, $C(GA;FE) =C(GO;DB) = - 1$, so $GFAE$ is harmonic, hence $P,E,F$ are collinear.

Let $O$ be the intersection of $BD$ and $AC$. The line parallel to $BC$ and passing through $P$ cuts $AC$ at $Q$. The line parallel to $CD$ and passing through $O$ cuts $PQ$ at $M$. $MO$ is the median of $\triangle BCO$. So it is the median of $\triangle PQO$ by $PQ \parallel BC$. Let $MO$ cut $AF$ at $R$. Because $MR \parallel CD$ and $AO = OC$, we have $AR=RF$ and $MR \perp AF$. So $MR$ bisects $AF$. This yields $MA=MF$. Since $\angle PAQ = 90^\circ $, we have $PM=MQ=MA=MF$. This concludes $M$ is the center of the circle passing through $P,Q,F$ and $A$. We have $\angle CAD = \angle BCA = \angle PQA$ because of parallel lines. We have $\angle EFA = \angle BCA$ because $ECFA$ is cyclic. We have $\angle PFA = \angle PQA$ because $PQFA$ is cyclic. So $\angle PFA =\angle EFA \Rightarrow P,F,A \text{ are collinear}$.

Let $Q$ be the intersection point of line $PA$ and $CD$, let $R$ be the intersection point of line $CE$ and $PA$. Let $\angle ACB = \alpha$ and $\angle DCA = \beta$. Let $O$ be the point of intersection of $AC$ and $BD$, i.e. the midpoint of $AC$. Let $r = OC = OA$. Note that $AE^{2} = RE*EC$ and $AF^{2} = QF*FC$. Hence, $ER = \frac{AE^{2}}{2r cos \alpha}$ and $FQ = \frac {FA^{2}}{2r cos \beta}$ (1) (Notice that triangle ACR is right angled and triangle ACQ is right angled) By Menelaus's Theorem on triangle $CQR$ and line $PDE$, we get $\frac{CB}{BR} \frac{RP}{PQ} \frac{QD}{DC} = 1$ Hence, $\frac {RP}{PQ} = \frac {BR}{CB} \frac{DC}{QD}$. We need to prove that $P,E,F$ are collinear, i.e. $\frac{CE}{ER} \frac{RP}{PQ} \frac{QF}{FC} = 1$, $\Leftrightarrow \frac{RP}{PQ} = \frac{ER}{EC} \frac{FC}{QF}$. $\Leftrightarrow \frac{ER}{EC} \frac{FC}{QF} = \frac {BR}{CB} \frac{DC}{QD}$ $\Leftrightarrow \frac{2r cos \beta * ER}{2r cos \alpha * FQ} = \frac {BR}{BC} * \frac{DC}{DQ}$. From (1), we get $\Leftrightarrow \frac{AE^{2}}{AF^{2}} \frac {cos^{2} \beta}{cos^{2} \alpha} = \frac{BR}{DQ} \frac{CD}{BC}$. (2) Note that $\frac{AE}{AF} = \frac {sin \alpha}{sin \beta}$. Hence (2) translates into proving: $\frac {tan^{2} \alpha}{tan^{2} \beta} = \frac {BR}{DQ} \frac{CD}{BC}$- (3) So let us prove (3). In triangle $ABR$ and triangle $ABC$, using sine law, we get $\frac {BR}{cos \beta} = \frac {AB}{sin (90 - \alpha)}$ (Note that $\angle ARB = 90 - \alpha$ as $\angle RAC = 90$) and $\frac {BC}{sin \beta} = \frac {AB}{sin \alpha}$ Hence, we get $\frac{BR}{BC} tan \beta = \frac{sin \alpha}{cos \alpha}$ i.e. $\frac{BR}{BC} = \frac{tan \alpha}{tan \beta}$. (*) Similarly, we get $\frac{CD}{DQ} = \frac{tan \alpha}{tan \beta}$. (**) Hence result follows by multiplying (*) and (**)

Eh...